Problem: Find the solution of the following equation whose argument is strictly between $90^\circ$ and $180^\circ$. Round your answer to the nearest thousandth. $z^3=27$ $z$ =
The Strategy A straightforward way to solve an equation of the form $z^{n}=m$ is by using the polar form of $z$. Therefore, our solution will consist of the following steps: Rewrite $z^n$ and $m$ in polar form. [How is this done, in general?] Solve for the modulus and argument of $z$. Find the rectangular form of $z$. [How is this done, in general?] Rewrite the equation in polar form Let's denote $r$ and $\theta$ to be the modulus and argument of $z$, respectively. Therefore, $z^{3}=r^{3}[\cos {({3}\cdot\theta)}+i\sin {({3}\cdot\theta)}]$. The number $27$ has a modulus of $27$. The argument of $27$ can be any multiple of $360^\circ$, so we can write it as $k\cdot360^\circ$ for an integer $k$. Now the equation looks as follows: $r^{3}[\cos {({3}\cdot\theta)}+i\sin {({3}\cdot\theta)}]=27[\cos(k\cdot360^\circ)+i\sin(k\cdot360^\circ)]$ When two complex numbers are equal, we know that both their moduli and arguments are equal. Therefore, we have the following equations for $r$ and $\theta$ : $r^{3}=27$ ${3}\cdot\theta=k\cdot 360^\circ$ Solving for $r$ $\begin{aligned}r^{3}&=27 \\\\ r &=3 \end{aligned}$ Solving for $\theta$ $\begin{aligned}{3}\cdot\theta&=k\cdot360^\circ \\\\\theta&=k\cdot120^\circ\end{aligned}$ Remember that $\theta$ is between $90^\circ$ and $180^\circ$. Therefore, we need to find the multiple of $120^\circ$ that is within that range. This multiple is simply $\theta=120^\circ$. Finding the rectangular form of $z$ Let's plug in $r=3$ and $\theta=120^\circ$ into the polar form of $z$ : $\begin{aligned}z&=r[\cos(\theta)+i\cdot\sin(\theta)]\\\\ &=3[\cos(120^\circ)+i\cdot\sin(120^\circ)]\\\\ &=3\cos(120^\circ)+3\sin(120^\circ)\cdot i\end{aligned}$ Using the calculator and rounding to the nearest thousandth, we get the following solution: $z=-1.5+2.598i$ Summary $z=-1.5+2.598i$